Question

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Answer 1

Answer:

6.00 moles of ScCl3 will be produced.

Explanation:

Step 1: Data given

Moles of Sc = 10.00 moles

Moles of Cl2 = 9.00 moles

Molar mass of Sc = 44.96 g/mol

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Sc + 3Cl2 → 2ScCl3

Step 3: Calculate the limiting reactant

For 2 moles Sc we need 3 moles Cl2 to produce 2 moles ScCl3

Cl2 is the limiting reactant. It will completely be consumed (9.00 moles)

Sc is the limiting reactant. There will react 2/3 * 9.00 = 6.00 moles

There will remain 10.00 - 6.00 =4.00 moles Sc

Step 4: Calculate moles ScCl3

For 3 moles Cl2 we'll have 2 moles ScCl3

For 9.00 moles we'll have 6.00 moles of ScCl3

6.00 moles of ScCl3 will be produced.

Answer 2

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$⇒$2ScCl_3$

The above reaction shows that 2 moles of Sc  can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$

$Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}$

Moles of ScCl_3= 6 moles